Question.
Does there exist a function $f(x,y)$ and a point $(x_0,y_0)$ such that:
- Both $f_{xy}$ and $f_{yx}$ exist in a neighbourhood of $(x_0,y_0)$.
- $f_{xy}(x_0, y_0)=f_{yx}(x_0, y_0)$.
- $f_{xy}$ is continuous at $(x_0,y_0)$.
- $f_{yx}$ is not continuous at $(x_0,y_0)$.
My thought.
Since there are functions $g(x,y)$ with $g_{xy}(0,0)=a\ne b= g_{yx}(0,0)$ (e.g. see this post), a possible way would be to construct a sequence of disjoint balls $B_n$ centered at $(1/n,1/n)$ with radius $1/(n+1)^3$, then smoothly restrict $g(x-1/n, y-1/n)$ in $B_n$ to get a function $$h_{n}(x,y)=\begin{cases} g(x-1/n,y-1/n) & \text{in a neighbourhood of}~ (1/n,1/n) \\ 0, &\text{outside}~B_n\end{cases}$$ and finally let $f(x,y)=\sum_{n\geq 1} h_n(x,y)$. In this way $$f_{xy}(1/n,1/n)=(h_n)_{xy}(1/n,1/n)\to a \quad\text{and}\quad f_{yx}(1/n,1/n)=(h_n)_{yx}(1/n,1/n)\to b $$ and the requirement 4 is met so long as $f_{xy}(0,0)=f_{yx}(0,0)=a$. But as to fulfill requirement 3, I cannot prove $f_{xy}(x_n,y_n)\to a$ when $(x_n,y_n)$ tends to $(0,0)$ arbitrarily, not just taking on points $(1/n,1/n)$.